Lesson # 14: Logarithm Equations & Identities

Today's lesson had two methods for solving equations, log= # and log=log

Method 1: Log = #                                                                                                                                             

Example:
log(2)x = 3                  1) Put into (b)^E = A form

2^3 = x

x = 8 

Example:

log(5)(x-3)+ log(5)x = log(5)10

log(5)x(x-3) = log(5)10               1) Put into B^e = A form

         x(x-3) = 10

       x^2- 3x = 10          2) Multiply both sides by x

  x^2-3x-10 = 0

   (x-5)(x+2) = 0        3) Simplify

           x = 5, -2     *-2 rejects*

           x = 5

*Note: negative answers are tricky, they must become positive in original equation or are rejected.*

       

Method 2: log = log                                                                                                                       

log5(x-3) + log5x = log5 10     1) Create only one log

log5(x-3)x = log5 10          2) Drop logs

(x-3)x = 10

x^2 - 3x = 10

x^2 -3x -10 = 0

(x-5)(x+2)

x= 5, -2            *-2 rejects*

x=5

Identities:

Equation        vs.         Identity

2x+3 = 5                    x + x = 2x

x=1                            x = all real numbers

x= specific #                rejects allowed

Prove the identity and state the value(s) of x for which it is true: (make both sides equal)

  

logx + log(x +3) = log(x^2 +3x)

  LEFT SIDE        RIGHT SIDE

Left side:  logx(x+3)

               logx^2 + 3x = Right side

*x must be great the 0     x>0*